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Inorganic Chemistry

Does anyone know the equation for the reaction between hydrogen-iodide and sulphuric acid to produce iodine sulphur and something else... .?

cheers.

Reply 1

Codefusion

8HI(g) + H₂SO4 → 4I₂(g) + H₂S(g) + 4H₂2O(l)
Iodine is a strong reducing agent , so it oxidises itself losing electrons to form Iodine and it reduces sulpher which form hydrogen sulhpide and the rest rect to form water.

Reply 2

firestone

According to my revision book the reaction goes,

2HI(g) + H₂SO₄(aq) → SO₂(g) + I₂(g) + 2H₂O(l)

Then further reduction to H₂S:
6HI + SO₂(g) → H₂S(g) + 3I₂(s) + 2H₂O(l)

Reply 3

Prometheus

There are three

NaI + H2SO4 --> HI + NaHSO4 then the HI goes onto reduce the SO42-
2HI + H2S04 --> SO2 + I2 + 2H2O then HI reduces the SO2
6HI + SO2 --> H2S + 3I2 + 2H2O

It is a progressive reduction
NaBr does only the first two (with Br instead of I)
NaCl does only the first one (with Cl instead of I)